Alexis wanted to understand whether grade level had any relationship to their opinion on extending the school day. She surveyed some students and displayed the results in the table below:

In favor Opposed Undecided
Grade 9 6 4 10
Grade 10 9 13 10
Grade 11 16 17 11
Grade 12 13 7 12

Compare P(Grade 10 | opposed) with P(opposed | Grade 10).
P(Grade 10 | opposed) = P(opposed | Grade 10)
P(Grade 10 | opposed) < P(opposed | Grade 10)
P(Grade 10| opposed) > P(opposed | Grade 10)
There is not enough information.

PLEASE someone help me. Thank you.

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barnuts barnuts · Answer 1 · 2017-08-25T15:20:55+02:00

Actually there is enough information to solve this problem. First, let us find the total per row and per column.

(see attached pic)

P(Grade 10 | opposed) with P(opposed | Grade 10)

P(Grade 10 | opposed) = Number in Grade 10 who are opposed / Total number of Opposed (column)

P(Grade 10 | opposed) = 13 / 41 = 0.3171

P(opposed | Grade 10) = Number in Grade 10 who are opposed / Total number in Grade 10 (row)

P(opposed | Grade 10) = 13 / 32 = 0.4063

Therefore:

P(Grade 10 | opposed) IS NOT EQUAL P(opposed | Grade 10), hence they are dependent events.

Answer:

P(Grade 10 | opposed) < P(opposed | Grade 10)

wagonbelleville wagonbelleville · Answer 2 · 2019-03-13T06:20:36+01:00

Answer:

B. P(Grade 10|Opposed) < P(Opposed|Grade 10)

Step-by-step explanation:

We are given,

The table representing the relation between grade level and extending school days.

The conditional probability of A given that B has occurred is,

[tex]P(A|B)=\frac{P(A\bigcap B)}{P(B)}[/tex].

So, we have,

[tex]P(Grade 10|Opposed)=\frac{P(Grade10\bigcap Opposed)}{P(Opposed)}[/tex]

i.e. [tex]P(Grade 10|Opposed)=\frac{13}{41}[/tex]

i.e. P(Grade 10|Opposed) = 0.32

Also, we have,

[tex]P(Opposed|Grade10)=\frac{P(Opposed\bigcap Grade10)}{P(Grade10)}[/tex]

i.e. [tex]P(Opposed|Grade10)=\frac{13}{32}[/tex]

i.e. P(Opposed|Grade10) = 0.41

Thus, we see that,

0.32 = P(Grade 10|Opposed) < P(Opposed|Grade 10) = 0.41

Hence, option B is correct.